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Home , Bijection, Injective function, Subset, Uncountable set, Union (set theory)

MATH 220 (all sections)—Homework #12 not to be turned in posted Friday, November 24, 2017

Definition: A A is finite if there exists a nonnegative c such that there exists a from A to {n ∈ N: n ≤ c}. (The integer c is called the of A.) I. (a) Let A be a finite set, and let B be a of A. Prove that B is finite. (Hint: induction on |A|. Note that our proof can’t use induction on |B|, or indeed refer to “the of elements in B” at all, because we don’t yet know that B is finite!) (b) Prove that the of two disjoint finite sets is finite. (c) Prove that the union of any two finite sets is finite. (Hint: A ∪ B = A ∪ (B − A).) (a) We proceed by induction on the nonnegative integer c in the definition that A is finite (the cardinality of c). step: c = 0. Then there is a bijection from A to {n ∈ N: n ≤ 0} = ∅, and thus A = ∅ (and, for that matter, the bijection is also the empty ). This forces B = ∅ as well (since that is the only subset of ∅), which is certainly finite. Induction step: Let c ≥ 0. The induction hypothesis is: if A has cardinality c, then any subset of A is finite. Now suppose A has cardinality c + 1, and let f : A → {n ∈ −1 N: n ≤ c + 1} be a bijection. Define A1 = f ({1, . . . , c}), and note that f restricted to A1 is a bijection from A1 to {1, . . . , c}, so that A1 has cardinality c. Finally, let B be a subset of A. −1 −1 Case 1: f (c + 1) ∈/ B. Then B is a subset of f ({1, . . . , c}) = A1, which has cardinality c; by the induction hypothesis, B is finite. −1 −1 Case 2: f (c + 1) ∈ B. Then define B1 = B − f (c + 1). As before, B1 is a subset of A1 and hence finite. Let d be the cardinality of B1 and let g1 : B1 → {n ∈ N: n ≤ d} be a bijection. Now define a function g : B → {n ∈ N: n ≤ d + 1} by ( g (x), if x ∈ B , g(x) = 1 1 d + 1, if x = f −1(c + 1). It is not hard to check that g is a bijection from B to {n ∈ N: n ≤ d + 1}. In particular, B is finite. (b) Let C and D be finite sets with C ∩ D = ∅, and let f : C → {n ∈ N: n ≤ c} and g : D → {n ∈ N: n ≤ d} be . Define h: C ∪ D → {n ∈ N: n ≤ c + d} by ( f(x), if x ∈ C, h(x) = g(x) + c, if x ∈ D. It is (somewhat tedious but) not hard to check that h is a bijection from C ∪ D to {n ∈ N: n ≤ c + d}. (The fact that C ∩ D = ∅ is necessary to show that h is a well-defined function.) In particular, C ∪ D is finite. (c) Let A and B be finite sets. Then B − A is a subset of the finite set B and hence is itself finite by part (a). Consequently, since A and B − A are always disjoint, the set A ∪ B = A ∪ (B − A) is the union of two disjoint finite sets and is therefore finite by part (b). II. Let A and B be nonempty sets. Prove that there exists an f : A → B there exists a g : B → A. First assume that f : A → B is injective. Let D = f(A) be the range of A; then f is a bijection from A to D. Choose any a ∈ A (possible since A is nonempty). Define g : B → A by ( f −1(y), if y ∈ D, g(y) = a, if y ∈ B − D. It is not hard to show that g is a well-defined function that is surjective. Conversely, assume that g : B → A is surjective. Define a function f : A → B as follows: for each a ∈ A, choose ba ∈ B such that g(ba) = a, and define f(a) = ba. It is not hard to check that f is a well-defined function that is injective. [Interested students who just read this proof might wish to find some introductory information about something called the “ of Choice”.]

SECTION 13.1 Show that the two given sets have equal cardinality by describing a bijection from one to the other. Describe√ your bijection with a formula (not as a table). 2. R and ( 2, ∞) 4. The set of even and the set of odd integers 8. Z and S = {x ∈ R: sin x = 1} 10. {0, 1} × N and Z 14. N × N and {(n, m) ∈ N × N: n ≤ m}. (Hint: draw “graphs” of both sets. The northwest–southeast diagonal slices of the first set look a lot like the horizontal slices of the second set. . . .) √ 2. Adapting√ one of the bijections from the book, define f : R → ( 2, ∞) by f(x) = x e + 2. It’s easy√ to check that f is a well-defined and bijective function (for example, f −1(y) = ln(y − 2) is its ). 4. A bijection f : {even integers} → {odd integers} is f(n) = n + 1; or f(n) = n − 1; or indeed f(n) = n + k or f(n) = n − k for any fixed odd integer k. 7π 3π π 5π 8. Note that S = {x ∈ R: sin x = 1} = {..., − 2 , − 2 , 2 , 2 ,... }. We see that one π bijection f : Z → S is f(n) = 2πn + 2 . 10. Define a function f : {0, 1} × N → Z by ( n, if b = 0, f(b, n) = 1 − n, if b = 1. It is not hard to show that f is a bijection; for example, its inverse function is ( (0, m), if m ≥ 1, f −1(m) = (1, 1 − m), if m ≤ 0. 14. Inspired by the given hint, define a function f : {(n, m) ∈ N × N: n ≤ m} → N × N by f(n, m) = (n, m + 1 − n). Since n ≤ m for elements of the domain, we see that m+1−n ≥ 1, and so the values really do lie in the . One can check that this function is a bijection; for example, its inverse function is f −1(x, y) = (x, x+y−1). III. Suppose that two sets A and B have the same cardinality. Prove that P(A) and P(B) have the same cardinality as each other.

Let f : A → B be a bijection. Define a function g : P(A) → P(B) by g(X) = f(X) for any X ∈ P(A). (Let’s interpret the notation carefully: g(X) is a function value since X is an of the domain P(A) of g, while f(X) is an since X is a subset of the domain A of f.) We claim that g is a bijection. First, let Y ∈ P(B). Then if we set X = f −1(Y ), then g(X) = f(X) = f(f −1(Y )) = Y . (The identity f(f −1(Y )) = Y isn’t true in general, but it is true when f is surjective, as shown on the previous homework.) In particular, g is surjective. Next, let X1 and X2 be elements of P(A) and suppose that g(X1) = g(X2). Then −1 −1 f(X1) = f(X2) (as images). Taking preimages of both sets yields f (f(X1)) = f (f(X2)). −1 −1 Since f is injective, we have f (f(X1)) = X1 and f (f(X2)) = X2 as shown on the previous homework; therefore X1 = X2. In particular, h is surjective.

SECTION 13.2 2. Prove that the set A = {(m, n) ∈ N × N: m ≤ n} is countably infinite. We already know that N×N is countably infinite. By problem 14 in Section 13.1 (above), A has the same cardinality as N × N. Therefore A is also countably infinite. 4. Prove that the set of all irrational is uncountable. Let I be the set of irrational numbers. Suppose for the sake of that I is countably infinite. We already know that Q is countably infinite. Then R = Q∪I, the union of two countably infinite sets, would be countably infinite as well; but this contradicts the known fact that R is uncountable. Therefore I is not countably infinite; since I is certainly an infinite set, we conclude that I is uncountable. 6. Prove or disprove: There exists a bijective function f : Q → R. Disproof: if there were such a bijective function, then Q and R would have the same cardinality. But we know that Q is countably infinite while R is uncountable, and therefore they do not have the same cardinality. We conclude that there is no bijection from Q to R. 8. Prove or disprove: The set Z × Q is countably infinite. Proof: we know that both Z and Q are countably infinite, and we know that the of two countably infinite sets is again countably infinite. Therefore Z × Q is countably infinite.

12. Describe a partition of N that divides N into ℵ0 countably infinite . We know that N × N is countably infinite, so let f : N × N → N be a bijection. For  each m ∈ N, define Am = f {m} × N . The sets {m} × N form a partition of the domain N × N; since f is a bijection, the sets Am form a partition of the codomain N. Also, each set {m} × N is countably infinite (there is an obvious bijection from each to N), and therefore their images Am under the bijection f are also each countably infinite. Therefore {Am : m ∈ N} is the desired partition of N; this partition has |N| = ℵ0 elements, as desired.  m−1 For example, if we take the bijection f : N×N → N defined by f (m, n) = 2 (2n− m−1 1), as in problem #15 of Section 13.2, then Am = {2 (2n − 1): n ∈ N} is the set of all positive integers that are divisible by 2m−1 but not divisible by 2m. For example,

A1 = {1, 3, 5, 7, 9, 11,... }

A2 = {2, 6, 10, 14, 18, 22,... }

A3 = {4, 12, 20, 28, 36, 44,... }

A4 = {8, 24, 40, 56, 72, 88,... } . .

These {Am : m ∈ N} form a partition of N into ℵ0 countably infinite subsets. 14. Suppose A = {(m, n) ∈ N × R: n = πm}. Is it true that |N| = |A|? Yes. Note that A = {(1, π), (2, 2π), (3, 3π),... }. Define a function f : A → N by  f (m, n) = m. It is easy to check that f is a bijection. Therefore |A| = |N|.

SECTION 13.3 4. Prove or disprove: If A ⊆ B ⊆ C and A and C are countably infinite, then B is countably infinite. Proof: Since A is infinite and A ⊆ B, we see that B is infinite. (This is the contrapositive of the first problem of this homework.) Then B is an infinite subset of the countably infinite set C, and therefore B is itself countably infinite. 6. Prove or disprove: Every infinite set is a subset of a countably infinite set. Disproof: consider R, which is uncountable. If R were a subset of a countably infinite set, then it too would be countably infinite, which is a contradiction. Therefore R is an infinite set that is not a subset of any countably infinite set. (Indeed, no is a subset of a countably infinite set.)

8. Prove or disprove: The set {(a1, a2, a3,... ): ai ∈ Z} of infinite of integers is countably infinite. Disproof: let S = {(a1, a2, a3,... ): ai ∈ Z}, and let f : N → S be a function. We claim that f is not surjective. In particular, this will show that there is no bijection from N to S, and so S is not countably infinite. (j) Let ai denote the ith element in the f(j). Define a sequence (b1, b2, b3,... ) where ( (i) 1, if ai 6= 1, bi = (i) −1, if ai = 1.

Then (b1, b2, b3,... ) ∈ S. On the other hand, for every i, we see that (b1, b2, b3,... ) 6= (i) f(i), since their ith coordinates bi and ai are different. Therefore (b1, b2, b3,... ) is not in the range of f, and therefore f is not surjective. 9. Prove that if A and B are finite sets with |A| = |B|, then any injection f : A → B is also a surjection. Show this is not necessarily true if A and B are not finite. Suppose, for the sake of contradiction, that f is not surjective, and choose b ∈ B that is not in the range of f. Let z be an object that is not an element of A, define A1 = A ∪ {z}, and define a function g : A1 → B by ( b, if x = z, g(x) = f(x), if x 6= z.

It is easy to check that g is also injective. However, |A1| = |A| + 1 = |B| + 1 > |B|, which violates the and is thus a contradiction. Therefore f must be surjective. If A and B are not finite, we have counterexamples such as A = N, B = Z, and f : A → B being the inclusion f(n) = n; then f is injective but not surjective (since −1 is not in the range, for example). 10. Prove that if A and B are finite sets with |A| = |B|, then any surjection f : A → B is also an injection. Show this is not necessarily true if A and B are not finite. Suppose, for the sake of contradiction, that f is not injective, and choose a1, a2 ∈ A with a1 6= a2 such that f(a1) = f(a2). Define A1 = A−{a2}, and define g : A1 → B by g(x) = f(x). It is easy to check that g is also surjective. However, |A1| = |A|−1 = |B|−1 < |B|, which violates the Pigeonhole Principle and is thus a contradiction. Therefore f must be injective. If A and B are not finite, we have counterexamples such as A = N × N, B = N, and f(m, n) = n; then f is surjective but not injective (since f(2, 3) = 3 = f(7, 3), for example).